[RC5] RC5 question

Bruce Wilson bruce.wilson at distributed.net
Fri Oct 3 21:58:32 EDT 2003


Cool numbers.  I can live with an initial estimate of 101 years.

What you also haven't factored in is the possibility of enlisting more
people than we currently have.  Of course, some may choose to work on
OGR or The Next Big Thing (whatever that may be).  Still, we can
exceed these estimates and finish sooner than your estimate by
attracting more people to participate.


__
Bruce Wilson <bwilson at distributed.net>
PGP KeyID: 5430B995, http://www.toomuchblue.com/ 

"I want to move to Theory. Everything works in Theory."
    --John Cash, id Software


| -----Original Message-----
| From: rc5-bounces at lists.distributed.net 
| [mailto:rc5-bounces at lists.distributed.net] On Behalf Of 
| Daniel Quintiliani
| Sent: Friday, October 03, 2003 20:33
| To: D.net Discussion
| Subject: Re: [RC5] RC5 question
| 
| 
| On Sat, 4 Oct 2003 09:05:04 +0800, Elektron 
| <elektron_rc5 at yahoo.ca> wrote:
| 
| >
| > Taking the latest RC5-64 numbers,
| > 10,428,970,063,364,096 Keys were completed yesterday 
| (0.056536% of the 
| > keyspace) at a sustained rate of 120,705,672,030 Keys/sec.
| >
| > That's 3809103098378452560 Keys/year (let's call it k). If 
| computing 
| > power doubles every 18 months (1.5 years), then a^1.5=2, or 
| a=2^(1/1.5) 
| > or approx. 1.587401051968. So our keyrate with respect to 
| time t (years) 
| > is ka^t, so our keys tested (big K) (by integration) is 
| ka^t ln(a) + C, 
| > and since we have no keys tested when t=0, k ln(a) + C = 0, 
| so K = ka^t 
| > ln(a) - k ln(a). K=2^128.
| >
| > 2^128 = k ln(a) (a^t-1)
| > 2^128/(k ln(a)) = a^t-1
| > 2^128/(k ln(a)) + 1 = a^t
| > t = ln(2^128/(k ln(a)))/ln(a) = 101.08435774324143022752
| >
| > Of course, I'm assuming RC5-128 is as fast as RC5-64 (which 
| it isn't), 
| > and that moore's law holds (which it might). Interestingly, 
| the last time 
| > I did this, I got 600 years. Somebody hit me if I did my 
| math wrong. But 
| > either way, 101 years is a lot more than my lifetime.
| >
| 
| Using yesterday's RC5-72 rate for a more accurate measure of time:
| 
| 114327804047 Keys/s * 60 * 60 * 24 * 365.25
| (goes upstairs and grabs TI-83+ from backpack)
| k = 3.607911127 * 10^8
| 
| Plugging it into your formula:
| 
| t = ln(2^128/((3.607911127*10^8) ln(2^(1/1.5)))) / ln(2^(1/1.5))
| 
| t = 101.201789 years
| 
| -- 
| -Dan
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